Proof that the fundamental group is actually a group/Proof
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Proof
We wish to show that the set [ilmath]\pi_1(X,b):=\frac{\Omega(X,b)}{\big({\small(\cdot)}\simeq{\small (\cdot)}\ (\text{rel }\{0,1\})\big)}[/ilmath] is actually a group with the operation [ilmath]\overline{*} [/ilmath] as described in the outline.
- Factoring:
- Setup:
- [ilmath]*:\Omega(X,b)\times\Omega(X,b)\rightarrow\Omega(X,b)[/ilmath] - the operation of loop concatenation - [ilmath]*:(\ell_1,\ell_2)\mapsto\left(\ell_1*\ell_2:I\rightarrow X\text{ by }\ell_1*\ell_2:t\mapsto\left\{\begin{array}{lr}\ell_1(2t) & \text{for }t\in[0,\frac{1}{2}]\\ \ell(2t-1) & \text{for }t\in[\frac{1}{2},1]\end{array}\right.\right)[/ilmath]
- through
- [ilmath](p,p):\Omega(X,b)\times\Omega(X,b)\rightarrow\pi_1(X,b)\times\pi_1(X,b)[/ilmath] by [ilmath](p,p):(\ell_1,\ell_2)\mapsto(p(\ell_1),p(\ell_2))[/ilmath]
- where [ilmath]p:\Omega(X,b)\rightarrow\pi_1(X,b)[/ilmath] is the canonical projection of the equivalence relation. As such we may say:
- [ilmath](p,p)[/ilmath] is given by by [ilmath](p,p):(\ell_1,\ell_2)\mapsto([\ell_1],[\ell_2])[/ilmath] instead
- We must show:
- [ilmath]\forall\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)\left[\big([\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2']\big)\implies\big([\ell_1*\ell_2]=[\ell_1'*\ell_2']\big)\right][/ilmath]^{[Note 1]}
- Proof:
- Let [ilmath]\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)[/ilmath] be given
- Suppose that [ilmath]\neg([\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'])[/ilmath] holds, then by the nature of logical implication we're done, as we do not care about the RHS's truth or falsity in this case
- Suppose that [ilmath][\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'][/ilmath] holds. We must show that in this case we have [ilmath][\ell_1*\ell_2]=[\ell_1'*\ell_2'][/ilmath]
- By homotopy invariance of loop concatenation we see exactly the desired result
- Since [ilmath]\ell_1,\ell_2,\ell_1'[/ilmath] and [ilmath]\ell_2'[/ilmath] were arbitrary this holds for all.
- Let [ilmath]\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)[/ilmath] be given
- Conclusion
- We obtain [ilmath]\overline{*}:\pi_1(X,b)\times\pi_1(X,b)\rightarrow\pi_1(X,b)[/ilmath] given unambiguously by:
- [ilmath]\overline{*}:([\ell_1],[\ell_2])\mapsto[\ell_1*\ell_2][/ilmath]
- We obtain [ilmath]\overline{*}:\pi_1(X,b)\times\pi_1(X,b)\rightarrow\pi_1(X,b)[/ilmath] given unambiguously by:
- Thus the group operation is:
- [ilmath][\ell_1]\overline{*}[\ell_2]:=[\ell_1*\ell_2][/ilmath]
- Setup:
- Associativity of the operation [ilmath]\overline{*} [/ilmath]
- Existence of an identity element in [ilmath](\pi_1(X,b),\overline{*})[/ilmath]
- For each element of [ilmath]\pi_1(X,b)[/ilmath] the existence of an inverse element in [ilmath](\pi_1(X,b),\overline{*})[/ilmath]
Grade: A*
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Notes
- ↑ Note that we turn [ilmath]([\ell_1],[\ell_2])=([\ell_1'],[\ell_2'])[/ilmath] into [ilmath][\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'][/ilmath] by using the defining property of an ordered pair
References